The temperature of a body falls from ${50^o}C$to ${40^o}C$ in $10$ minutes. If the temperature of the surroundings is ${20^o}C$ Then temperature of the body after another $10$ minutes will be ........ $^oC$
Diffcult
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(b) In first case $\frac{{50 - 40}}{{10}} = K\left[ {\frac{{50 + 40}}{2} - 20} \right]$….$(i)$
In second case $\frac{{40 - {\theta _2}}}{{10}} = K\left[ {\frac{{40 + {\theta _2}}}{2} - 20} \right]$….$(ii)$
By solving ${\theta _2} = {33.3^o}C$.
In second case $\frac{{40 - {\theta _2}}}{{10}} = K\left[ {\frac{{40 + {\theta _2}}}{2} - 20} \right]$….$(ii)$
By solving ${\theta _2} = {33.3^o}C$.
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